Question

Integration Questions.
$∫ (6^6^6^x) × (6^6^x) × (6^x)dx$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle \int {6}^{ {6}^{ {6}^{x} } } {6}^{ {6}^{x} } {6}^{x} \: dx$

To solve this integral, we use method of Substitution.

So, Substitute

$\rm :\longmapsto\: {6}^{ {6}^{x} } = y$

So,

$\rm :\longmapsto\: \dfrac{d}{dx}{6}^{ {6}^{x} } =\dfrac{d}{dx} y$

We know,

$\boxed{\tt{ \dfrac{d}{dx} {a}^{x} = {a}^{x} \: loga \: }}$

So, using this, we get

$\rm :\longmapsto\: {6}^{ {6}^{x} } \: log6 \: \dfrac{d}{dx} {6}^{x} = \dfrac{dy}{dx}$

$\rm :\longmapsto\: {6}^{ {6}^{x} } \: log6 \: {6}^{x} \: log6 = \dfrac{dy}{dx}$

$\rm :\longmapsto\: {6}^{ {6}^{x} } \: {6}^{x} \: {(log6)}^{2} = \dfrac{dy}{dx}$

$\rm\implies \: {6}^{ {6}^{x} } {6}^{x} \: dx \: = \dfrac{dy}{ {(log6)}^{2} }$

So, on substituting the values in above integral, we get

$\rm \:  =  \: \displaystyle \int {6}^{y} \dfrac{dy}{ {(log6)}^{2} }$

$\rm \:  =  \: \dfrac{1}{ {(log6)}^{2} } \displaystyle \int {6}^{y} \: dy$

$\rm \:  =  \: \dfrac{1}{ {(log6)}^{2} } \times \dfrac{ {6}^{y} }{log6} + c$

$\rm \:  =  \: \dfrac{ {6}^{y} }{ {(log6)}^{3} } + c$

$\rm \:  =  \: \dfrac{ {6}^{ {6}^{ {6}^{x} } } }{ {(log6)}^{3} } + c$

Hence,

$\rm\implies \:\boxed{\tt{ \: \: \displaystyle \int {6}^{ {6}^{ {6}^{x} } } \: dx \: = \: \frac{ {6}^{ {6}^{ {6}^{x} } } }{ {(log6)}^{3}} + c \: \: }}$

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More to Know

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle \int {6}^{ {6}^{ {6}^{x} } } {6}^{ {6}^{x} } {6}^{x} \: dx$

To solve this integral, we use method of Substitution.

So, Substitute

$\rm :\longmapsto\: {6}^{ {6}^{x} } = y$

So,

$\rm :\longmapsto\: \dfrac{d}{dx}{6}^{ {6}^{x} } =\dfrac{d}{dx} y$

We know,

$\boxed{\tt{ \dfrac{d}{dx} {a}^{x} = {a}^{x} \: loga \: }}$

So, using this, we get

$\rm :\longmapsto\: {6}^{ {6}^{x} } \: log6 \: \dfrac{d}{dx} {6}^{x} = \dfrac{dy}{dx}$

$\rm :\longmapsto\: {6}^{ {6}^{x} } \: log6 \: {6}^{x} \: log6 = \dfrac{dy}{dx}$

$\rm :\longmapsto\: {6}^{ {6}^{x} } \: {6}^{x} \: {(log6)}^{2} = \dfrac{dy}{dx}$

$\rm\implies \: {6}^{ {6}^{x} } {6}^{x} \: dx \: = \dfrac{dy}{ {(log6)}^{2} }$

So, on substituting the values in above integral, we get

$\rm \:  =  \: \displaystyle \int {6}^{y} \dfrac{dy}{ {(log6)}^{2} }$

$\rm \:  =  \: \dfrac{1}{ {(log6)}^{2} } \displaystyle \int {6}^{y} \: dy$

$\rm \:  =  \: \dfrac{1}{ {(log6)}^{2} } \times \dfrac{ {6}^{y} }{log6} + c$

$\rm \:  =  \: \dfrac{ {6}^{y} }{ {(log6)}^{3} } + c$

$\rm \:  =  \: \dfrac{ {6}^{ {6}^{ {6}^{x} } } }{ {(log6)}^{3} } + c$

Hence,

$\rm\implies \:\boxed{\tt{ \: \: \displaystyle \int {6}^{ {6}^{ {6}^{x} } } \: dx \: = \: \frac{ {6}^{ {6}^{ {6}^{x} } } }{ {(log6)}^{3}} + c \: \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to Know

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$